Chapter 10 Rotation

10.2 The Rotational Variables A rigid body is a body that can rotate with all its parts locked together and without any change in its shape. A fixed axis means that the rotation occurs about an axis that does not move. Figure skater Sasha Cohen in motion of pure rotation about a vertical axis. (Elsa/Getty Images, Inc.)

10.2 The Rotational Variables

An angle defined in this way is measured in radians (rad). 10.2 The Rotational Variables: Angular Position Here s is the length of a circular arc that extends from the x axis (the zero angular position) to the reference line, and r is the radius of the circle. An angle defined in this way is measured in radians (rad).

angular displacement Dq given by 10.2 The Rotational Variables: Angular Displacement If a body rotates about the rotation axis as in, changing the angular position of the reference line from q1 to q2, the body undergoes an angular displacement Dq given by An angular displacement in the counterclockwise direction is positive, and one in the clockwise direction is negative.

10.2 The Rotational Variables: Angular Velocity Suppose that our rotating body is at angular position q1 at time t1 and at angular position q2 at time t2. Then the average angular velocity of the body in the time interval t from t1 to t2 to be. The instantaneous angular velocity w is the limit of the ratio as Dt approaches zero.

10.2 The Rotational Variables: Angular Acceleration If the angular velocity of a rotating body is not constant, then the body has an angular acceleration. If w2 and w1 be its angular velocities at times t2 and t1, respectively, then the average angular acceleration of the rotating body in the interval from t1 to t2 is defined as The instantaneous angular acceleration a, is the limit of this quantity as Dt approaches zero. These relations hold for every particle of that body. The unit of angular acceleration is (rad/s2).

Sample problem

Sample problem

Sample problem

Sample problem, Angular Velocity and Acceleration

10.3: Are Angular Quantities Vectors?

10.4: Rotation with Constant Angular Acceleration Just as in the basic equations for constant linear acceleration, the basic equations for constant angular acceleration can be derived in a similar manner. The constant angular acceleration equations are similar to the constant linear acceleration equations.

Sample problem: Constant Angular Acceleration (b) Describe the grindstone’s rotation between t =0 and t =32 s. Description: The wheel is initially rotating in the negative (clockwise) direction with angular velocity w0=4.6 rad/s, but its angular acceleration a is positive. The initial opposite signs of angular velocity and angular acceleration means that the wheel slows in its rotation in the negative direction, stops, and then reverses to rotate in the positive direction. After the reference line comes back through its initial orientation of q= 0, the wheel turns an additional 5.0 rev by time t =32 s. (c) At what time t does the grindstone momentarily stop? Calculation: With w =0, we solve for the corresponding time t. The angular acceleration is constant, so we can use the rotation equation: Substituting known values and setting q0 =0 and q =5.0 rev =10p rad give us Solving this quadratic equation for t, we find t =32 s.

Sample problem: Constant Angular Acceleration

10.5: Relating Linear and Angular Variables If a reference line on a rigid body rotates through an angle q, a point within the body at a position r from the rotation axis moves a distance s along a circular arc, where s is given by: Differentiating the above equation with respect to time—with r held constant—leads to The period of revolution T for the motion of each point and for the rigid body itself is given by Substituting for v we find also that

10.5: Relating Linear and Angular Variables Differentiating the velocity relation with respect to time—again with r held constant— leads to Here, a =dw/dt. Note that dv/dt =at represents only the part of the linear acceleration that is responsible for changes in the magnitude v of the linear velocity. Like v, that part of the linear acceleration is tangent to the path of the point in question. Also, the radial part of the acceleration is the centripetal acceleration given by

Sample problem Consider an induction roller coaster (which can be accelerated by magnetic forces even on a horizontal track). Each passenger is to leave the loading point with acceleration g along the horizontal track. That first section of track forms a circular arc (Fig. 10-10), so that the passenger also experiences a centripetal acceleration. As the passenger accelerates along the arc, the magnitude of this centripetal acceleration increases alarmingly. When the magnitude a of the net acceleration reaches 4g at some point P and angle qP along the arc, the passenger moves in a straight line, along a tangent to the arc. What angle qP should the arc subtend so that a is 4g at point P? Calculations: Substituting wo=0, and qo=0, and we find: But which gives: This leads us to a total acceleration: Substituting for ar, and solving for q lead to: When a reaches the design value of 4g, angle q is the angle qP . Substituting a =4g, q= qP, and at= g, we find:

Sample problem, cont. (b) What is the magnitude a of the passenger’s net acceleration at point P and after point P? Reasoning: At P, a has the design value of 4g. Just after P is reached, the passenger moves in a straight line and no longer has centripetal acceleration. Thus, the passenger has only the acceleration magnitude g along the track. Hence, a =4g at P and a =g after P. (Answer) Roller-coaster headache can occur when a passenger’s head undergoes an abrupt change in acceleration, with the acceleration magnitude large before or after the change. The reason is that the change can cause the brain to move relative to the skull, tearing the veins that bridge the brain and skull. Our design to increase the acceleration from g to 4g along the path to P might harm the passenger, but the abrupt change in acceleration as the passenger passes through point P is more likely to cause roller-coaster headache.

10.6: Kinetic Energy of Rotation For an extended rotating rigid body, treat the body as a collection of particles with different speeds, and add up the kinetic energies of all the particles to find the total kinetic energy of the body: (mi is the mass of the ith particle and vi is its speed). (w is the same for all particles). The quantity in parentheses on the right side is called the rotational inertia (or moment of inertia) I of the body with respect to the axis of rotation. It is a constant for a particular rigid body and a particular rotation axis. (That axis must always be specified.) Therefore,

10.7: Calculating the Rotational Inertia If a rigid body consists of a great many adjacent particles (it is continuous, like a Frisbee), we consider an integral and define the rotational inertia of the body as

Parallel Axis Theorem: 10.7: Calculating the Rotational Inertia Parallel Axis Theorem: If h is a perpendicular distance between a given axis and the axis through the center of mass (these two axes being parallel).Then the rotational inertia I about the given axis is Let O be the center of mass (and also the origin of the coordinate system) of the arbitrarily shaped body shown in cross section. Consider an axis through O perpendicular to the plane of the figure, and another axis through point P parallel to the first axis. Let the x and y coordinates of P be a and b. Let dm be a mass element with the general coordinates x and y. The rotational inertia of the body about the axis through P is: But x2 + y2 =R2, where R is the distance from O to dm, the first integral is simply Icom, the rotational inertia of the body about an axis through its center of mass. The last term in is Mh2, where M is the body’s total mass.

Sample problem: Rotational Inertia

Sample problem: Rotational Inertia

Sample problem: Rotational Inertia

Sample problem: Rotational KE

10.8: Torque The ability of a force F to rotate the body depends on both the magnitude of its tangential component Ft, and also on just how far from O, the pivot point, the force is applied. To include both these factors, a quantity called torque t is defined as: OR, where is called the moment arm of F.

10.9: Newton’s Law of Rotation For more than one force, we can generalize:

Sample problem: Newton’s 2nd Law in Rotational Motion The torque from the tension force, T, is -RT, negative because the torque rotates the disk clockwise from rest. The rotational inertia I of the disk is ½ MR2. But tnet =Ia =-RT=1/2 MR2a. Because the cord does not slip, the linear acceleration a of the block and the (tangential) linear acceleration at of the rim of the disk are equal. We now have: T=-1/2 Ma. Combining results: We then find T: The angular acceleration of the disk is: Note that the acceleration a of the falling block is less than g, and tension T in the cord (=6.0 N) is less than the gravitational force on the hanging block ( mg =11.8 N). Sample problem: Newton’s 2nd Law in Rotational Motion Forces on block: From the block’s freebody, we can write Newton’s second law for components along a vertical y axis as: T –mg= ma.

10.10: Work and Rotational Kinetic Energy where t is the torque doing the work W, and qi and qf are the body’s angular positions before and after the work is done, respectively. When t is constant, The rate at which the work is done is the power

10.10: Work and Rotational Kinetic Energy

Sample problem: Work, Rotational KE, Torque